An object of mass 4.40 kg is projected into the air at a 55 degre angle. it hits the ground 3.8 s later. what is its change in momentum while it is in the air? ignore air resistance.
Impulse = M*g*(3.8 s), downward.
The launch angle and velocity are not needed for the answer, but they do determine when the object hits the ground. .
that doesn't work as the correct answer :/
The launch velocity is 22.73 m/s, with a vertical component of 18.6 m/s. It his the top of the trajectory at 18.6/g = 1.90 s and hits the ground at 3.80 s.
If you use my suggested method, you get
momentum change = 4.40 kg*9.8 m/s^2*3.8s
= 163.9 kg m/s
Initial vertical momentum =
4.4*sin55*22.73 = 81/93 kg*m/s
That gets double on the way dwn, for a net momentum change of 163.9 kg m/s
That answer should work. Your course's "correct" answer may be wrong.
To calculate the change in momentum of an object, we first need to determine its initial and final momentum.
The initial momentum can be found using the equation:
p_initial = m * v_initial
Where:
- p_initial is the initial momentum
- m is the mass of the object
- v_initial is the initial velocity of the object
We are given the mass of the object, which is 4.40 kg. However, the initial velocity is not directly provided. Instead, we have the launch angle and the time it takes to hit the ground.
To find the initial velocity, we need to break it down into its horizontal and vertical components. Given the launch angle of 55 degrees, we can use trigonometry to find the initial vertical velocity (v_y_initial) and the initial horizontal velocity (v_x_initial):
v_y_initial = v_initial * sin(angle)
v_x_initial = v_initial * cos(angle)
Where:
- v_y_initial is the initial vertical velocity
- v_x_initial is the initial horizontal velocity
- v_initial is the initial velocity
- angle is the launch angle
Since the only component affected by gravity is the vertical component, we can calculate the time it takes for the object to reach its highest point using the equation:
t_highestPoint = v_y_initial / g
Where:
- t_highestPoint is the time to reach the highest point
- v_y_initial is the initial vertical velocity
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
Given the time it takes to hit the ground (3.8 s), we can calculate the time the object takes to reach its highest point (t_highestPoint). The total time of flight, which is the time the object is in the air, is twice the time to reach the highest point:
t_flight = 2 * t_highestPoint
Next, we can substitute the time of flight and the equation of motion in the vertical direction:
h = v_y_initial * t_flight - (1/2) * g * t_flight^2
Where:
- h is the maximum height reached by the object
Since the object returns to the ground, the maximum height reached is zero. By substituting this into the equation, we can solve for the initial vertical velocity (v_y_initial):
0 = v_y_initial * t_flight - (1/2) * g * t_flight^2
Solving for v_y_initial gives us:
v_y_initial = (1/2) * g * t_flight
Finally, we can calculate the initial velocity (v_initial) by using the Pythagorean theorem:
v_initial = sqrt(v_y_initial^2 + v_x_initial^2)
Now that we have the initial velocity, we can calculate the initial momentum (p_initial):
p_initial = m * v_initial
After calculating the initial momentum, we need to find the final momentum (p_final), which will be inversely proportional to the initial momentum since the mass does not change.
p_final = -p_initial
The negative sign indicates that the direction of the momentum is reversed upon hitting the ground.
Finally, we can calculate the change in momentum:
Δp = p_final - p_initial
Substituting the values, you can now calculate the change in momentum of the object while it is in the air.